Introductory calculus

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

The formula for the derivative of a function $f(x)$ is

So the correct formula is Option C.

Simplify the expression $f(x + h) = 3(x + h)^{2}$.

The question tells us that

Substituting in $(x + h)$, we get

So the correct expression is Option A.

Use your answer from Question 2.

We know the expression for $f(x + h)$ from Question 2, so we need to subtract $f(x)$ and simplify.

Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

Now we have enough information to look at the entire derivation:

The final thing that we need to do before evaluating the limit $h \to 0$ is to cancel the $h$ in the numerator and denominator. Cancelling the $h$ is crucial, because as long as $h$ is in the denominator, substituting $h = 0$ will produce an undefined result. The cancellation of $h$ allows us to evaluate the limit (substitute $h = 0$).

So the correct choice is Option D.

All that's left to do is to evaluate the limit $h \to 0$.

The final step is to evaluate the limit $h \to 0$

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

The formula for the derivative of a function $f(x)$ is

So the correct formula is Option C.

Simplify the expression $f(x + h) = 7(x + h)^{2}$.

The question tells us that

Substituting in $(x + h)$, we get

So the correct expression is Option C.

Use your answer from Question 2.

We know the expression for $f(x + h)$ from Question 2, so we need to subtract $f(x)$ and simplify.

Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

Now we have enough information to look at the entire derivation:

The final thing that we need to do before evaluating the limit $h \to 0$ is to cancel the $h$ in the numerator and denominator. Cancelling the $h$ is crucial, because as long as $h$ is in the denominator, substituting $h = 0$ will produce an undefined result. The cancellation of $h$ allows us to evaluate the limit (substitute $h = 0$).

So the correct choice is Option B.

All that's left to do is to evaluate the limit $h \to 0$.

The final step is to evaluate the limit $h \to 0$

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

The formula for the derivative of a function $f(x)$ is

So the correct formula is Option B.

Simplify the expression $f(x + h) = 6(x + h)^{2}$.

The question tells us that

Substituting in $(x + h)$, we get

So the correct expression is Option D.

Use your answer from Question 2.

We know the expression for $f(x + h)$ from Question 2, so we need to subtract $f(x)$ and simplify.

Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

Now we have enough information to look at the entire derivation:

The final thing that we need to do before evaluating the limit $h \to 0$ is to cancel the $h$ in the numerator and denominator. Cancelling the $h$ is crucial, because as long as $h$ is in the denominator, substituting $h = 0$ will produce an undefined result. The cancellation of $h$ allows us to evaluate the limit (substitute $h = 0$).

So the correct choice is Option A.

All that's left to do is to evaluate the limit $h \to 0$.

The final step is to evaluate the limit $h \to 0$

The "rate of change" means finding the derivative of the function for a particular value of $d$. Therefore we need to use the derivative of the function to calculate the answer to this question.

The question tells us an equation which describes a real situation: there is water in a lake and the amount of water is changing. There is a pump pushing water into or out of the lake as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the lake.

A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

Now use the derivative to find the rate of change at $d = 3$ days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when $d = 3$.

After 3 days, the rate of change of volume of the water in the lake is 11 kilolitres per day.

The rate of change tells us if the amount of water is increasing or decreasing.

The question asks us if the amount of water in the lake (the volume) is increasing or decreasing when $d = 3$ days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

The correct choice is: increasing because the rate of change is positive.

Do we use the volume function or the derivative to find the answer to this question?

We need to calculate the maximum volume according to the function $V(d)$. The question tells us that the volume is at its maximum value when $d = \frac{35}{8}$ days; and the volume formula tells us the volume of water in the lake at time $d$. To find the maximum volume, substitute $d = \frac{35}{8}$ into the function and evaluate.

The calculation shows that after $\frac{35}{8}$ days, the lake has a maximum volume of $\frac{1369}{16}$ kilolitres of water.

Therefore, the lake has a maximum volume of $\frac{1369}{16}$ ($\approx \text{85.56}$) kilolitres of water.

The "rate of change" means finding the derivative of the function for a particular value of $d$. Therefore we need to use the derivative of the function to calculate the answer to this question.

The question tells us an equation which describes a real situation: there is water in a lake and the amount of water is changing. There is a pump pushing water into or out of the lake as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the lake.

A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

Now use the derivative to find the rate of change at $d = 7$ days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when $d = 7$.

After 7 days, the rate of change of volume of the water in the lake is -23 kilolitres per day.

The rate of change tells us if the amount of water is increasing or decreasing.

The question asks us if the amount of water in the lake (the volume) is increasing or decreasing when $d = 7$ days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

The correct choice is: decreasing because the rate of change is negative.

Do we use the volume function or the derivative to find the answer to this question?

We need to calculate the maximum volume according to the function $V(d)$. The question tells us that the volume is at its maximum value when $d = \frac{47}{10}$ days; and the volume formula tells us the volume of water in the lake at time $d$. To find the maximum volume, substitute $d = \frac{47}{10}$ into the function and evaluate.

The calculation shows that after $\frac{47}{10}$ days, the lake has a maximum volume of $\frac{2809}{20}$ kilolitres of water.

Therefore, the lake has a maximum volume of $\frac{2809}{20}$ ($\approx \text{140.45}$) kilolitres of water.

The "rate of change" means finding the derivative of the function for a particular value of $d$. Therefore we need to use the derivative of the function to calculate the answer to this question.

The question tells us an equation which describes a real situation: there is water in a pond and the amount of water is changing. There is a pump pushing water into or out of the pond as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the pond.

A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

Now use the derivative to find the rate of change at $d = 4$ days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when $d = 4$.

After 4 days, the rate of change of volume of the water in the pond is 7 kilolitres per day.

The rate of change tells us if the amount of water is increasing or decreasing.

The question asks us if the amount of water in the pond (the volume) is increasing or decreasing when $d = 4$ days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

The correct choice is: increasing because the rate of change is positive.

Do we use the volume function or the derivative to find the answer to this question?

We must figure out when the pond will be empty. The words "the pond is empty" means that the volume is zero. Therefore, we should substitute $V = 0$ into the volume function and then solve for $d$. (The solution below uses factorisation, but we could also use the quadratic formula.)

We are calculating when the pond is empty: the answer cannot be negative because time (the number of days) cannot be negative. Therefore we should take only the positive answer.

Therefore, the pond will be empty after $\text{11}$ days.

Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find $h$ in terms of $r$:

Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of $r$.

The formula for the volume of a cone is

Since we want to find the derivative of $V$ with respect to $r$, we need to have the entire formula in terms of $r$.

We can now differentiate the expression for $V$ with respect to $r$:

We can now evaluate this at $r = \text{6}\ \text{cm}$.

Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find $h$ in terms of $r$:

Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of $r$.

The formula for the volume of a cone is

Since we want to find the derivative of $V$ with respect to $r$, we need to have the entire formula in terms of $r$.

We can now differentiate the expression for $V$ with respect to $r$:

We can now evaluate this at $r = \text{7}\ \text{cm}$.

Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find $r$ in terms of $h$:

Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of $h$.

The formula for the volume of a cone is

Since we want to find the derivative of $V$ with respect to $h$, we need to have the entire formula in terms of $h$.

We can now differentiate the expression for $V$ with respect to $h$:

We can now evaluate this at $h = \text{7}\ \text{cm}$.

To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: $r = \frac{1}{2} (3 y) = \frac{3 y}{2}$.

Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable $y$. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

It is common to move the quadratic terms of $y$ to the beginning of the function; so the shaded area is: $\; A(y) = - \frac{9 \pi}{4} y^{2} - 2 y^{2} + 33 y$.

Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

The function $\; A(y) = - \frac{9 \pi}{4} y^{2} - 2 y^{2} + 33 y = -(2 + \frac{9 \pi}{4})y^{2} + 33 y \; \;$ is a quadratic function. Notice that the quadratic coefficient, $\; -(2 + \frac{9 \pi}{4})$, is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

The question asks us to figure out which $y$ value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
We find the derivative of the function:

We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

The value of $y$ which maximizes the shaded area (rounded to 3 decimal places) is $\text{1.819}\ \text{cm}$. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

We know that the height of the fake money is $3 y$. We also know from question (2) that the shaded area will be maximised when $y = \text{1.819}$. Therefore, we need to substitute this value into the expression for the height:

When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be $\text{5.457}\ \text{cm}$.

To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: $r = \frac{1}{2} (3 y) = \frac{3 y}{2}$.

Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable $y$. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

It is common to move the quadratic terms of $y$ to the beginning of the function; so the shaded area is: $\; A(y) = - \frac{9 \pi}{4} y^{2} - 2 y^{2} + 18 y$.

Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

The function $\; A(y) = - \frac{9 \pi}{4} y^{2} - 2 y^{2} + 18 y = -(2 + \frac{9 \pi}{4})y^{2} + 18 y \; \;$ is a quadratic function. Notice that the quadratic coefficient, $\; -(2 + \frac{9 \pi}{4})$, is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

The question asks us to figure out which $y$ value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
We find the derivative of the function:

We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

The value of $y$ which maximizes the shaded area (rounded to 3 decimal places) is $\text{0.992}\ \text{cm}$. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

We know that the height of the fake money is $3 y$. We also know from question (2) that the shaded area will be maximised when $y = \text{0.992}$. Therefore, we need to substitute this value into the expression for the height:

When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be $\text{2.976}\ \text{cm}$.

To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: $r = \frac{1}{2} (5 x) = \frac{5 x}{2}$.

Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable $x$. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

It is common to move the quadratic terms of $x$ to the beginning of the function; so the shaded area is: $\; A(x) = - \frac{25 \pi}{4} x^{2} - 2 x^{2} + 30 x$.

Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

The function $\; A(x) = - \frac{25 \pi}{4} x^{2} - 2 x^{2} + 30 x = -(2 + \frac{25 \pi}{4})x^{2} + 30 x \; \;$ is a quadratic function. Notice that the quadratic coefficient, $\; -(2 + \frac{25 \pi}{4})$, is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

The question asks us to figure out which $x$ value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
We find the derivative of the function:

We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

The value of $x$ which maximizes the shaded area (rounded to 2 decimal places) is $\text{0.69}\ \text{cm}$. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

We know that the height of the fake money is $5 x$. We also know from question (2) that the shaded area will be maximised when $x = \text{0.69}$. Therefore, we need to substitute this value into the expression for the height:

When the shaded area is maximised, the height of the fake money (rounded to 2 decimal places) will be $\text{3.45}\ \text{cm}$.

The question asks you to write down the volume "in terms of $x$ and $h$." This means that you need to write an equation for the volume which contains both variables.

The volume of the box is the product of the three sides, in this case, $6 x$ $\text{cm}$, $4 x$ $\text{cm}$, and $h$ $\text{cm}$.

Therefore, the volume function for the box is $24 h x^{2}$.

Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

Therefore, the surface area equation for the box is $20 h x + 48 x^{2}$.

This question is similar to the previous one, except that now the equation we cannot include $h$. There is another piece of information we have not used yet: the volume of the box is $44$ $\text{cm$^{3}$}$. What can you do with that information?

This question requires that we write an equation for the surface area which does not include $h$. We have two equations which describe the box: $V = 24 h x^{2}$ and surface area $= 20 h x + 48 x^{2}$. We use the first of these equations to remove the $h$ from the second equation. Remember that the volume of the box must be 44 $\text{cm$^{3}$}$. Therefore, the equation for volume can be written as: $44 = 24 h x^{2}$.

Solve for $h$:

We have found that $h = \frac{11}{6 x^{2}}$.

Now we substitute this expression for $h$ into the equation for the surface area:

We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of $x$ is $\; 48 x^{2} + \frac{110}{3 x}$.

If you want to calculate the minimum surface area, you need to find the value of $x$ which minimises the function. To find a maximum or minimum of a function, use the derivative.

We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function $S.A.(x) = 48 x^{2} + \frac{110}{3 x}$.

It looks like the minimum surface area is close to 75 or 76. To find the minimum value, we need to find the derivative of the function:

Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
Solve for $x$:

The value of $x$ which gives the minimum surface area is $x_{\text{min}} \approx \text{0.73}$. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

We have found the $x$ value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating $S.A.(x_{\text{min}})$.

The minimum surface area for the box is $\text{75.8}\ \text{cm$^{2}$}$. This answer also agrees with the graph above.

Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of $x_{\text{min}}$ just as we did to find the minimum surface area. We already found that the height is $h = \frac{11}{6 x^{2}}$ and we also know that the width is simply $w = 6x$. Use these two formulae to calculate the minimum values for each quantity:

The height and width corresponding to the minimum surface area are $h =$ $\text{3.48}\ \text{cm}$ and $w =$ $\text{4.35}\ \text{cm}$.

Think about the cost for one of the boxes: the box has a surface area of $\text{75.8}\ \text{cm$^{2}$}$ and the cardboard used to make the box costs $\text{N}\llap{=}\,\text{0.004}$ per $\text{cm$^{2}$}$. How can you use these numbers to find the cost of one box?

We need to work out how much it will cost to make $\text{1,700}$ boxes if each box requires $\text{75.8}\ \text{cm$^{2}$}$ of cardboard. Given that that material for the boxes costs $\text{N}\llap{=}\,\text{0.004}$ per $\text{cm$^{2}$}$, the total cost will be:

Therefore, the total cost for $\text{1,700}$ boxes is $\text{N}\llap{=}\,\text{515.44}$.

The question asks you to write down the volume "in terms of $x$ and $h$." This means that you need to write an equation for the volume which contains both variables.

The volume of the box is the product of the three sides, in this case, $5 x$ $\text{cm}$, $2 x$ $\text{cm}$, and $h$ $\text{cm}$.

Therefore, the volume function for the box is $10 h x^{2}$.

Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

Therefore, the surface area equation for the box is $14 h x + 20 x^{2}$.

This question is similar to the previous one, except that now the equation we cannot include $h$. There is another piece of information we have not used yet: the volume of the box is $48$ $\text{cm$^{3}$}$. What can you do with that information?

This question requires that we write an equation for the surface area which does not include $h$. We have two equations which describe the box: $V = 10 h x^{2}$ and surface area $= 14 h x + 20 x^{2}$. We use the first of these equations to remove the $h$ from the second equation. Remember that the volume of the box must be 48 $\text{cm$^{3}$}$. Therefore, the equation for volume can be written as: $48 = 10 h x^{2}$.

Solve for $h$:

We have found that $h = \frac{24}{5 x^{2}}$.

Now we substitute this expression for $h$ into the equation for the surface area:

We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of $x$ is $\; 20 x^{2} + \frac{336}{5 x}$.

If you want to calculate the minimum surface area, you need to find the value of $x$ which minimises the function. To find a maximum or minimum of a function, use the derivative.

We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function $S.A.(x) = 20 x^{2} + \frac{336}{5 x}$.

It looks like the minimum surface area is close to 84 or 85. To find the minimum value, we need to find the derivative of the function:

Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
Solve for $x$:

The value of $x$ which gives the minimum surface area is $x_{\text{min}} \approx \text{1.19}$. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

We have found the $x$ value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating $S.A.(x_{\text{min}})$.

The minimum surface area for the box is $\text{84.79}\ \text{cm$^{2}$}$. This answer also agrees with the graph above.

Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of $x_{\text{min}}$ just as we did to find the minimum surface area. We already found that the height is $h = \frac{24}{5 x^{2}}$ and we also know that the width is simply $w = 5x$. Use these two formulae to calculate the minimum values for each quantity:

The height and width corresponding to the minimum surface area are $h =$ $\text{3.4}\ \text{cm}$ and $w =$ $\text{5.94}\ \text{cm}$.

Think about the cost for one of the boxes: the box has a surface area of $\text{84.79}\ \text{cm$^{2}$}$ and the cardboard used to make the box costs $\text{N}\llap{=}\,\text{0.006}$ per $\text{cm$^{2}$}$. How can you use these numbers to find the cost of one box?

We need to work out how much it will cost to make $\text{1,800}$ boxes if each box requires $\text{84.79}\ \text{cm$^{2}$}$ of cardboard. Given that that material for the boxes costs $\text{N}\llap{=}\,\text{0.006}$ per $\text{cm$^{2}$}$, the total cost will be:

Therefore, the total cost for $\text{1,800}$ boxes is $\text{N}\llap{=}\,\text{915.73}$.

The question asks you to write down the volume "in terms of $x$ and $h$." This means that you need to write an equation for the volume which contains both variables.

The volume of the box is the product of the three sides, in this case, $6 x$ $\text{cm}$, $3 x$ $\text{cm}$, and $h$ $\text{cm}$.

Therefore, the volume function for the box is $18 h x^{2}$.

Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

Therefore, the surface area equation for the box is $18 h x + 36 x^{2}$.

This question is similar to the previous one, except that now the equation we cannot include $h$. There is another piece of information we have not used yet: the volume of the box is $46$ $\text{cm$^{3}$}$. What can you do with that information?

This question requires that we write an equation for the surface area which does not include $h$. We have two equations which describe the box: $V = 18 h x^{2}$ and surface area $= 18 h x + 36 x^{2}$. We use the first of these equations to remove the $h$ from the second equation. Remember that the volume of the box must be 46 $\text{cm$^{3}$}$. Therefore, the equation for volume can be written as: $46 = 18 h x^{2}$.

Solve for $h$:

We have found that $h = \frac{23}{9 x^{2}}$.

Now we substitute this expression for $h$ into the equation for the surface area:

We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of $x$ is $\; 36 x^{2} + \frac{46}{x}$.

If you want to calculate the minimum surface area, you need to find the value of $x$ which minimises the function. To find a maximum or minimum of a function, use the derivative.

We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function $S.A.(x) = 36 x^{2} + \frac{46}{x}$.

It looks like the minimum surface area is close to 80 or 81. To find the minimum value, we need to find the derivative of the function:

Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
Solve for $x$:

The value of $x$ which gives the minimum surface area is $x_{\text{min}} \approx \text{0.86}$. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

We have found the $x$ value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating $S.A.(x_{\text{min}})$.

The minimum surface area for the box is $\text{80.11}\ \text{cm$^{2}$}$. This answer also agrees with the graph above.

Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of $x_{\text{min}}$ just as we did to find the minimum surface area. We already found that the height is $h = \frac{23}{9 x^{2}}$ and we also know that the width is simply $w = 6x$. Use these two formulae to calculate the minimum values for each quantity:

The height and width corresponding to the minimum surface area are $h =$ $\text{3.45}\ \text{cm}$ and $w =$ $\text{5.17}\ \text{cm}$.

Think about the cost for one of the boxes: the box has a surface area of $\text{80.11}\ \text{cm$^{2}$}$ and the cardboard used to make the box costs $\text{N}\llap{=}\,\text{0.006}$ per $\text{cm$^{2}$}$. How can you use these numbers to find the cost of one box?

We need to work out how much it will cost to make $\text{2,000}$ boxes if each box requires $\text{80.11}\ \text{cm$^{2}$}$ of cardboard. Given that that material for the boxes costs $\text{N}\llap{=}\,\text{0.006}$ per $\text{cm$^{2}$}$, the total cost will be:

Therefore, the total cost for $\text{2,000}$ boxes is $\text{N}\llap{=}\,\text{961.32}$.

Substitute $t$ with the number of minutes, $\text{3}$, and evaluate.

Substitute $t$ with the number of minutes, $\text{3}$, and evaluate.

The number of molecules of the medicine in the bloodstream $\text{3}$ minutes after the medicine was taken is $\text{252}$.

Calculate the derivative of the function $M(t)$ and then use it to answer this question.

To get the rate of change of the function $M(t)$, we need to calculate its derivative at a particular value of $t$. We will start by calculating the derivative.

We now need to substitute the given value of time to find the rate of change at that time

The rate at which the number of molecules of the medicine in the bloodstream is changing exactly $\text{4}$ minutes after the medicine was taken is $\text{75}$ molecules per minute.

You need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

We need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

We will calculate the second derivative from the first derivative which we obtained in Question 1.

We now need to equate the second derivative to zero and then solve for $t$.

The maximum rate of change of the number of molecules of the medicine in the bloodstream is after $\text{2}$ minutes.

Substitute $t$ with the number of minutes, $\text{5}$, and evaluate.

Substitute $t$ with the number of minutes, $\text{5}$, and evaluate.

The number of molecules of the medicine in the bloodstream $\text{5}$ minutes after the medicine was taken is $\text{235}$.

Calculate the derivative of the function $M(t)$ and then use it to answer this question.

To get the rate of change of the function $M(t)$, we need to calculate its derivative at a particular value of $t$. We will start by calculating the derivative.

We now need to substitute the given value of time to find the rate of change at that time

The rate at which the number of molecules of the medicine in the bloodstream is changing exactly $\text{2}$ minutes after the medicine was taken is $\text{58}$ molecules per minute.

You need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

We need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

We will calculate the second derivative from the first derivative which we obtained in Question 1.

We now need to equate the second derivative to zero and then solve for $t$.

The maximum rate of change of the number of molecules of the medicine in the bloodstream is after $\dfrac{\text{2}}{\text{3}}$ minutes.

Substitute $t$ with the number of minutes, $\text{2}$, and evaluate.

Substitute $t$ with the number of minutes, $\text{2}$, and evaluate.

The number of molecules of the drug in the bloodstream $\text{2}$ minutes after the drug was taken is $\text{118}$.

Calculate the derivative of the function $M(t)$ and then use it to answer this question.

To get the rate of change of the function $M(t)$, we need to calculate its derivative at a particular value of $t$. We will start by calculating the derivative.

We now need to substitute the given value of time to find the rate of change at that time

The rate at which the number of molecules of the drug in the bloodstream is changing exactly $\text{3}$ minutes after the drug was taken is $\text{36}$ molecules per minute.

You need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.

We need to start by calculating the second derivative of the function $M(t)$. In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.

We will calculate the second derivative from the first derivative which we obtained in Question 1.

We now need to equate the second derivative to zero and then solve for $t$.

The maximum rate of change of the number of molecules of the drug in the bloodstream is after $\text{0}$ minutes.